# Continuous bijections between $\mathbb{R}^2$ and $\mathbb{R}^2 \setminus \{(0,0)\}$

It is well-known that $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^2 \setminus \{(0,0)\}$.

I have two questions.

a) Does there exist a continuous bijection $f: \mathbb{R}^2 \to \mathbb{R}^2 \setminus \{(0,0)\}$ ?

b) Does there exist a continuous bijection $g: \mathbb{R}^2 \setminus \{(0,0)\} \to \mathbb{R}^2$ ?

• You should notice that a) implies b): make $g=f^{-1}$ – Carlos Eugenio Thompson Pinzón Nov 9 '13 at 1:31
• As Tim Kimsella said, such a continuous function would be a homeomorphism. But it can't be a homeomorphism because $X:= \mathbb{R} ^2 - \left\{ 0\right\}$ and $Y := \mathbb{R} ^2$ have different homotopy types ( $X\equiv S^1$ and $Y\equiv \ast$ ). The argument would be the same for $\phi : X\to Y$ or $\varphi : Y\to X$. – Fernando Jan 1 '15 at 5:16